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高中 2022 级第二次诊断性考试
数学参考答案及评分标准
一、选择题：本题共8小题，每小题5分，共40分．
1．D 2．A 3．C 4．B 5．D 6．B 7．A 8．C
二、选择题：本大题共3小题，每小题6分，共18分．在每小题给出的四个选项中，有多项符合题目
要求．全部选对的得6分，选对但不全的得部分分，有选错的得0分．
9．BCD 10．BC 11．ABD
三、填空题：本题共3个小题，每小题5分，共15分．
12．3； 13． ； 14．
四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17分，共77分．解
答应写出文字说明、证明过程或演算步骤．
5．解：（1）∵ ，
1
由正弦定理得：
，····························································2分
又 则 ，
，
∴ ，····················································································4分
∴
，又B是三角形内角，··········································································5分
∴
；········································································································6分
2）∵ ，且 ，
（
∴
，···································································································8分
∴
，·························································································9分
∴
，·························································································11分
∴
．····································································································13分
16．解：（1）a=0时， ， ，且 ，······························2分
∴ ，·································································································4分
故切线方程为：y−(e+1)=e(x−1)，即ex−y+1=0；·························································6分
（2）∵ ， ，·········································································7分
由1<a<e，存在 ，使得 ，即 ， ，································9分
当 时， ， 单调递减；························································10分
当 时， ， 单调递增，·························································11分
故 ，··············································12分
令 ， ，················································13分
∴ 在 上单调递减，··············································································14分
数学参考答案 第1页（共6页）易知 ，所以 ．·········································································15分
数学参考答案 第2页（共6页）17．解：（1）设数列 的公差为d，
令 ，得 ，所以 ，····································································2分
令 ，得 ，所以 ，·························································4分
∵数列 的公差大于0，
∴ ， ，所以 ；·······································································6分
（2）（i）k=1时，n=1，则 ；········································································7分
k=1时， ，··································································8分
∴ ， ；·················································································9分
（ii）由题意可知： ，························································10分
①当 时， ，则 ，∴ ；···················································11分
②当 时， ，则 ，∴ ；·······································12分
③当 时， ，则 ，∴ ，········································13分
∴
，················································································14分
∴ ．·······························································15分
数学参考答案 第3页（共6页）18．解法一：（1）证明：如图1，设AC的中点为F，连接BF并延长交CD于点G，
易知△ABC为等边三角形，且BG⊥AC，······································································1分
在△ACD中，AC=AD=2，满足 ，
∴AD⊥AC，则G为CD中点，···················································································2分
又点E为PD中点，G为CD中点，
∴EG//PC，又PC⊥AC，
∴AC⊥EG，··········································································································3分
而BG，EG均在平面BEG内，且BG∩EG=G，
∴AC⊥平面BEG， ，
∴AC⊥BE；···········································································································5分
（2）因为AC⊥平面BEF，则AC⊥EF，AC⊥FG，··························································6分
因此∠EFG即为二面角E-AC-D的平面角，
∴ ，而 ，···············································································7分
∴在△EFG中， ，由余弦定理得 ，································8分
∴ ，则EG⊥FG，············································································9分
∴EG⊥平面ABCD，PC⊥平面ABCD， ， ，
由AD⊥AC，则△ACD的面积为2，···········································································10分
∴三棱锥P-ACD的体积 ；··························································11分
（3）由PC//EG，则直线PC与平面ACE所成角即为直线EG与平面ACE所成角，
由（1）可知AC⊥平面BEG， ，
∴平面BEG⊥平面ACE，因此∠FEG即为所求角，·····················································12分
思路一：在△EFG中， ， ，由正弦定理： ，············13分
∴ ，·······················································································14分
当 ，即如（2）中EG⊥平面ABCD时，·····················································15分
，···················································································16分
∴直线PC与平面ACE所成角的正弦值为 ．··························································17分
思路二：在△EFG中， ， ，
故点G在以F为圆心，1为半径的圆上，···································································12分
当EG与该圆相切时，即如（2）中EG⊥平面ABCD时，∠FEG最大，····························14分
此时 ，·····················································································16分
数学参考答案 第4页（共6页）∴直线PC与平面ACE所成角的正弦值为 ．··························································17分
解法二：（1）证明：易知△ABC为等边三角形在△ACD中，AC=AD=2，
满足 ，∴AD⊥AC，········································································1分
以A为坐标原点，建立上如图所示的空间直角坐标系A-xyz，
设点P(x，y，z)，C(2，0，0)，D(0，2，0)，B(1， ，0)，E( ， ， )，
易知 ， ， ···························2分
由PC⊥AC，则2(2−x)=0，则x=2，·············································································3分
则 ·························································································4分
∴ ，则AC⊥BE；······················································································5分
（2）由（1）可知AC⊥平面BEG， ，
∴平面BEG⊥平面ABCD，·······················································································6分
以A为坐标原点，建立如图所示的空间直角坐标系A-xyz，
设点E(1，m，n)（m<2，n>0），易知 ， ，
设平面ACE的法向量为：n=(x，y，z)，
1
∴ ，不妨令 ，则平面ACE的一个法向量为：n=(0，−n，m)，
1
又平面ACD的一个法向量为n=(0，0，1)，·································································7分
2
∴cos<n，n>= ，······························································8分
1 2
∵ ，则 ，
解得：n= ，则点E到平面ACD的距离为 ，
由E为PD的中点，则点P到平面ACD的距离为 ，·················································9分
在△ACD中，AC=AD=2，满足 ，
∴AD⊥AC，则△ACD的面积为2，···········································································10分
∴三棱锥P-ACD的体积 ；··························································11分
（3）由PC//EG，则直线PC与平面ACE所成角即为直线EG与平面ACE所成角，设为θ，
由 ，则 ，·································································12分
·····················13分
∴
······························14分
∴
（当且仅当m=1时，等号成立）·······················16分
数学参考答案 第5页（共6页）即 的最大值为 ，
∴直线PC与平面ACE所成角正弦值的最大值为 ．················································17分
19．解：（1）由已知得， ，即 ，······················································1分
又离心率为 ，则 ，
∵ ，所以 ，即 ，·····························································2分
∴ ， ，·································································································3分
∴椭圆Г的标准方程为： ；······································································4分
（2）设点M(x，y)，则点M满足： ，则 ，
0 0
由已知可得F(0， )，H(0， )，设直线MF与MH的斜率分别为 ， ，
∴ ， ，···············································································5分
直线MF与MH的斜率之积满足： ．······················6分
（i）∵D(2， )，G(2，0)，则 ， ，
直线PF的方程为： ，令 ，则 ，
∴ ，·····································································································7分
直线HQ的方程为： ，令 ，可得Q(2， )，
∴ ，
∴ ，················································································8分
且 ，·························································9分
∴ ；···················································································10分
(ii) 存在 ，使得|TK|定值 ，理由如下：···················································11分
设点 ， ， ，
①当过椭圆上点 的直线l斜率存在时，设直线l方程为： ，
数学参考答案 第6页（共6页）带入椭圆Г的方程： ，
化简并整理得： ，
∵直线l与椭圆仅有一个公共点，
∴ ，
化简得： ，····················································································12分
∴ ，代入 ，得 ，
∴ ，
从而直线l的方程为： ，即 ，·············································13分
②当过 的直线l斜率不存在且与椭圆Г仅有一个公共点时，直线l的方程为：
满足上式．
同理：当过椭圆上点 的直线： 与椭圆Г仅有一个公共点，
这两条直线都过点S，所以有 ， ，
∴直线IJ的方程为 ．···········································································14分
由（i）则直线PF的方程为： ，令 ，则R( ， )，
又Q(2， )，
∴RQ的中点S( ， )，即 ， ．
∴直线IJ的方程表示为： ，
即 ，·······································································15分
令 ，解得： ，·······························································16分
∴直线IJ恒过定点N( )，又∵OT⊥IJ，
∴点T在以ON为直径的圆上，即K( )，|TK|定值 ．·······································17分
数学参考答案 第7页（共6页）
